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MCF3M Functions & Applications
Unit 1 — Introduction to the Quadratic Function
Domain and Range of a Quadratic Function

MCF3M Functions & Applications · Unit 1 — Introduction to the Quadratic Function

Lesson 1.7

Domain and Range of a Quadratic Function

Trace parabolas with the Domain & Range Scanner, use the vertex and opening to state the range, and see why real-world models restrict domain and range.

Estimated time: 30–45 minutes

Lesson progress0%

Learning goals

determine the domain and range of a quadratic from its graph or equation
use the vertex to find the minimum or maximum
connect opening direction to the range inequality
distinguish mathematical and contextual domain and range

Step 1 of 4

You can revisit any step at any time — your answers stay as you left them.

Step 1 of 4

Warm Up

Real-world idea

How far can it go?

Imagine extending this parabola forever. Where could x go? How low could y ever get?

y(x) = (x + 1)² − 4

Warm Up

The parabola above opens up and has vertex (−1, −4). What is the lowest y-value it ever reaches?

Your answer

Warm Up

True or false: if we kept drawing this parabola, x could be any real number.

TrueFalse

Step 2 of 4

Learn

Part 1 — Domain: x can go anywhere

You can put ANY real number into a(x − h)² + k and get an output. So an unrestricted quadratic has domain {x ∈ ℝ}: all real numbers.

Part 2 — Range: the vertex sets the limit

The range depends on two things: the vertex's y-value k, and which way the parabola opens.

Range from vertex form

a > 0 → opens up → minimum k → {y ∈ ℝ | y ≥ k}
a < 0 → opens down → maximum k → {y ∈ ℝ | y ≤ k}

≥ not >

The vertex is ON the graph, so y = k is reached. Use ≥ or ≤ — never > or <.

Part 3 — The Domain & Range Scanner

Trace along each parabola. Watch the arrows, the vertex and the dashed range boundary through the vertex. Then answer the scanner's questions.

Domain & Range Scanner

g(x) = 2(x + 3)² − 5

Trace along the curve: move x

(−1, 3)

Vertex: (−3, −5) — minimum
Axis of symmetry: x = −3
Opens: up

Can x continue farther left?

Can x continue farther right?

Can y go below −5?

Domain?

Range?

Try it

What is the range of y = −4(x − 1)² + 9?

a{y ∈ ℝ | y ≥ 9}b{y ∈ ℝ | y ≤ 9}c{y ∈ ℝ | y ≤ 1}d{y ∈ ℝ}

Try it

Which statement about y = 5(x + 2)² − 7 is true?

aMaximum value −7bMinimum value −7cMinimum value −2dMaximum value 5

Key term

Contextual domain and range

The values of the inputs and outputs that make sense in the real situation. They are often a smaller part of the mathematical domain and range.

Try it

In the fountain model h(d) = −0.5(d − 2)² + 2, d is the horizontal distance (m) from the nozzle and h is the water's height (m).

Why is d = 6 not part of the contextual domain?

aBecause h(6) cannot be calculated.bBecause the water lands at d = 4; at d = 6 the formula gives h = −6, which is below the ground, so it doesn't describe real water.cBecause 6 is larger than the vertex.dIt is part of the contextual domain.

Worked example

State the domain and range of y = −3(x + 4)² − 1. Is −1 a maximum or a minimum?

Step 1
Domain: no restriction.
{x ∈ ℝ}
Step 2
Vertex (−4, −1); a < 0, so it opens down.
Step 3
Highest y is −1, and it is reached.
{y ∈ ℝ | y ≤ −1}

Answer: D: {x ∈ ℝ}; R: {y ∈ ℝ | y ≤ −1}; −1 is a maximum.

Step 3 of 4

Practice

Build it

Vertex + opening → range.

Multiple choice

A parabola has vertex (3, −2) and opens up. What is its range?

a{y ∈ ℝ | y ≥ −2}b{y ∈ ℝ | y ≤ −2}c{y ∈ ℝ | y ≥ 3}d{y ∈ ℝ | y > −2}

Multiple choice

A parabola has vertex (−1, 5) and opens down. What is its maximum value?

a−1b5cNo maximumd−5

Apply it

Domain and range from vertex form.

Multiple choice

What are the domain and range of y = 0.25(x + 6)² + 2?

aD: {x ∈ ℝ}; R: {y ∈ ℝ | y ≥ 2}bD: {x ∈ ℝ | x ≥ −6}; R: {y ∈ ℝ | y ≥ 2}cD: {x ∈ ℝ}; R: {y ∈ ℝ | y ≤ 2}dD: {x ∈ ℝ}; R: {y ∈ ℝ | y ≥ −6}

Multiple choice

What is the range of y = −3(x − 4)²?

a{y ∈ ℝ | y ≤ 4}b{y ∈ ℝ | y ≥ 0}c{y ∈ ℝ | y ≤ 0}d{y ∈ ℝ | y ≤ −3}

Challenge

Restricted domain and range in context.

Multiple choice

A paper airplane is launched from the floor. Its height is h(t) = −0.5(t − 3)² + 4.5 metres after t seconds. It lands at t = 6.

What is the contextual domain?

a{t ∈ ℝ}b{t ∈ ℝ | 0 ≤ t ≤ 6}c{t ∈ ℝ | t ≥ 3}d{t ∈ ℝ | 0 ≤ t ≤ 4.5}

Multiple choice

Same paper airplane: h(t) = −0.5(t − 3)² + 4.5, for 0 ≤ t ≤ 6.

What is the contextual range?

a{h ∈ ℝ | h ≤ 4.5}b{h ∈ ℝ | 0 ≤ h ≤ 4.5}c{h ∈ ℝ | 0 ≤ h ≤ 6}d{h ∈ ℝ | h ≥ 0}

Explain it

Multiple choice

A student says: “Every quadratic has domain all real numbers, so every real-world quadratic model should use all real numbers too.” Why is this incomplete?

aQuadratics don't have domain all real numbers.bThe function itself accepts any real x, but a model is restricted to values that make sense — e.g., no negative time and no height below the ground.cReal-world models never use quadratics.dThe student is completely correct.

True or false

True or false: the range of y = 2(x − 1)² + 3 is {y ∈ ℝ | y > 3}.

TrueFalse

Want more practice? (1 optional questions)

Extra practice

Which parabola has range {y ∈ ℝ | y ≤ −3}?

ay = (x − 3)² − 3by = −(x + 1)² − 3cy = −(x − 3)² + 3dy = 2(x + 3)²

Step 4 of 4

Mastery Check

Lesson 1.7 mastery check

6 questions · no hints during the check. Answer them all, then check your score — you can retry as many times as you like.

Question 1 of 6
What is the domain of y = −(x + 5)² + 2?
{x ∈ ℝ | x ≤ −5}{x ∈ ℝ}{x ∈ ℝ | x ≤ 2}{x ∈ ℝ | x ≥ −5}
Question 2 of 6
What is the range of y = 3(x − 4)² − 1?
{y ∈ ℝ | y ≥ −1}{y ∈ ℝ | y ≤ −1}{y ∈ ℝ | y ≥ 4}{y ∈ ℝ | y > −1}
Question 3 of 6
What is the range of y = −0.5(x + 2)² + 6?
{y ∈ ℝ | y ≥ 6}{y ∈ ℝ | y ≤ 6}{y ∈ ℝ | y ≤ −2}{y ∈ ℝ}
Question 4 of 6
Which is true for y = −2(x − 3)² + 8?
Minimum 8 at x = 3Maximum 8 at x = 3Maximum 3 at x = 8Minimum −2
Question 5 of 6
True or false: the range of y = x² − 9 is {y ∈ ℝ | y ≤ −9}.
YesNo
Question 6 of 6
A soccer ball is kicked from the ground. Its height is h(t) = −5(t − 1)² + 5 metres after t seconds, and it lands at t = 2.
What are the contextual domain and range?
D: {t ∈ ℝ}; R: {h ∈ ℝ | h ≤ 5}D: {t ∈ ℝ | 0 ≤ t ≤ 2}; R: {h ∈ ℝ | 0 ≤ h ≤ 5}D: {t ∈ ℝ | 0 ≤ t ≤ 5}; R: {h ∈ ℝ | 0 ≤ h ≤ 2}D: {t ∈ ℝ | t ≥ 0}; R: {h ∈ ℝ | h ≥ 0}

Answer every question to check your work.

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Curriculum content shown is temporary sample material for demonstration. Official Ontario curriculum content will replace it.